C3: Multiple Factors Flashcards
(10 cards)
Example One (P and A):
If we purchase a new car and it has a three-year warranty, then we should not have to pay any repair expenses during the first three years of ownership.
However, if we keep the car for 7 years, and if we have repair bills of $500 per year beginning in year 4 and continuing each year through year 7, then what is the present worth of those repairs if the interest rate is 5% per year?
i = 5%
A= $500 for years 4 thru 7
with n = 4.
P1 = the P value for A that will occur in year 3 based on the equation.
P = ? (the final P value in year 0 after moving the P1 value in year 3 back to year 0 by treating P1 as a F)
P=$500(P/A, 5%, 4)(P/F, 5%, 3)
P=$500 (3.5460) (0.8638)
P = $1,531.52 is the present value in year 0 of the repair bills that occur in years 4 through 7.
Example Two (A and A):
If we purchase a new car and it has a three-year warranty, then we should not have to pay any repair expenses during the first three years of ownership.
However, if we keep the car for 7 years, and if we have repair bills of $600 per year beginning in year 4 and continuing each year through year 7, then what is the annual value in years 1 through 7 of those repairs if the interest rate is 5% per year?
i = 5%.
A1 = $600 for years 4 thru 7
with n = 4.
F = the F value for A that will occur in year 7 based on the eqn.
A = ? (the final A value in years 1 thru 7)
A=$600(F/A, 5%, 4)(A/F, 5%, 7)
A = $600 (4.3101) (0.12282)
A = $317.62 is the annual value in years 1 through 7 of the repair bills that occur in years 4 through 7.
Example Three (P and G):
If we purchase a new car and it has a three-year warranty, then we should not have to pay any repair expenses during the first three years of ownership.
However, if we keep the car for 7 years, and if we have repair bills of $300 per year beginning in year 4 and the repairs bills continue to increase by $300 each year through year 7, then what is the present value in year 0 of those repairs if the interest rate is 5% per year?
i = 5%.
A1 = $600 for years 4 thru 7
with n = 4.
F = the F value for A that will occur in year 7 based on the equation.
A = ? (the final A value in years 1 through 7)
A=$600(F/A, 5%, 4)(A/F, 5%, 7)
A = $600 (4.3101) (0.12282)
A = $317.62 is the annual value in years 1 through 7 of the repair bills that occur in years 4 through 7.
3.1 –
Toyota is evaluating a robot that will automatically remove molded parts from a die molding machine. If the robot is purchased, then production costs should decrease by $50,000 per year in each of the first 3 years and after that the production costs should decrease by $75,000 per year in years 4 and 5 due to an anticipated increase in sales and output.
If the interest rate is 15%, calculate the present worth of the cost savings of this automation project.
A1=$50k for n1=3
A2=$75k for n2=2,
i = 15%, P = ?
P = $50k (P/A, 15%, 3)
+ $75k (P/A,15%, 2)(P/F, 15%, 3)
P = $50k (2.2832)
+ $75k (1.6257)(0.6575)
P = $194,327.33
3.2 –
Best Value Store’s gross revenue was 65% of its total revenue over the most recent 4-year period. If the total revenue was $380,000,000 for years 1 and 2, and the total revenue was $420,000,000 for years 3 and 4, and the interest rate was 20% per year, calculate the equivalent annual worth of the gross revenue over the entire 4-year period.
n1 = 2, A1 = (0.65) ($380M) A1 = $247M
n2 = 2, A2 = (0.65) ($420M) A2 = $273M
n = 2 + 2 = 4,
i = 20%, A = ?
A=$247M·(P/A, 20%,2)(A/P, 20%,4)
+ $273M·(F/A, 20%, 2)(A/F, 20%, 4)
A = $247,000,000 (1.5278) (0.38629) + $273,000,000 (2.2000) (0.18629)
A = $145,772,943.90 + $111,885,774.00 **A = $257,658,717.90**
3.3 –
A regional utility supplier generates peak-demand power using diesel-powered generators. The company is considering switching to natural gas but it will take time to make the transition. After the transition is complete the company anticipates that it will save $24,000 per year, with the first savings beginning 2 years from now. If the interest rate is 8% per year, calculate the equivalent annual worth of this project for the next ten years (years 1 through 10).
A1 = $24,000 for years 2 to 10
with n1 = 9, i = 8%, n =10, A = ?
A=$24k (F/A, 8%,9)(A/F, 8%,10)
A=$24k (12.4876) (0.06903)
A=$20,688.46
3.4 –
A bank savings account contains $30,000 today. The account has been earning interest at the rate of 7% per year for the past 15 years. The last deposit into the savings account was made 10 years ago. Prior to that an equal amount of money was deposited each year for a total of 5 years for a total of 5 deposits during the past 15 years. How much money was deposited each year during that 5 year time period?
It will be necessary to view today as being 15 years in the future.
F = $30,000 with n =15, i = 7%, **A = ?** with n = 5
A=$30k (P/F, 7%, 15)(A/P, 7%, 5)
A=$30k (0.3624)(0.24389)
A=$2,651.57
3.5 –
Refer to the following cash flow diagram.
If the interest rate is 12% per year, calculate the amount of money in year 5 that would be equivalent to all the cash flows shown from years 0 through 13.
i = 12%
A1 = $4,000 with n1 = 4 A2 = $8,000 with n2 = 7
P=$4k (F/A, 12%,4)(F/P, 12%,2)
+ $8k (F/A, 12%,7)(P/F, 12%,8)
P=$4k (4.7793)(1.2544)
+ $8k (10.0890)(0.4039)
P =$23,980.62
+ $32,599.58
P = $56,580.20
3.6 –
The state of Colorado is considering a new dam to provide electricity for its projected population growth over the next 50-years. The dam is expected to generate electricity that will provide revenue of $2,700,000 per year for the next 4 years. However, simulation projections indicate that rainfall will begin to decrease in year 5 and this will result in a continuing decrease in electricity and in revenue by $40,000 per year beginning in year 5 and every year thereafter (negative gradient). If the interest rate is 6% per year, and the state of Colorado wishes the dam to pay for itself in 25 years, how much money could be invested in the dam project today?
A = $2,700,000
for n = 25 years
G = -$40,000 from years
5 thru 25 with n = 22
(n = number of Gs + 1)
i = 6%, P = ?
P = $2.7M (P/A, 6%, 25)
– $40k(P/G, 6%,22)(P/F, 6%,3)
P = $2,7M (12.7834)
- $40k (98.9412)(0.8396)
- *P = $31,192,338.74**
